Question

How many 8-bits sequences that start with the same two bits or their fourth and fifth bits are equal or end with the same two bits are there?

Answer 1

Answer on Question #83760 – Math – Discrete Mathematics

Question

How many 8-bits sequences that start with the same two bits or their fourth and fifth bits are equal or end with the same two bits are there?

Solution

The total number of 8-bit sequences is $2^8 = 256$.

Denote $N_{1}$ the number of sequences with the stated property. We have

$$N_{1} = 256 - N_{2}$$

where $N_{2}$ is the number of 8-bit sequences with all different pairs: $1^{\text{st}}$ and $2^{\text{nd}}$ different bits, and $4^{\text{th}}$ and $5^{\text{th}}$ different bits, and $7^{\text{th}}$ and $8^{\text{th}}$ different bits.

Calculate $N_{2}$.

There are 2 ways to choose the $1^{\text{st}}$ bit, after that the $2^{\text{nd}}$ bit is defined automatically – it is opposite to the $1^{\text{st}}$. There are 2 ways to choose the $3^{\text{rd}}$ bit, 2 ways to choose the $4^{\text{th}}$ bit – $5^{\text{th}}$ is defined automatically, 2 ways to choose the $6^{\text{th}}$ bit, 2 ways to choose the $7^{\text{th}}$ bit – $8^{\text{th}}$ is defined automatically.

So totally there are $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$ ways, $N_{2} = 32$.

Then $N_{1} = 256 - N_{2} = 224$.

Answer: 224.

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